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I am learning Relations and I was confused by the fact that simple assignment relations

The given relation $R= \{(x, y) | x-y = 0 \} \subset \mathbb{R} \times \mathbb{R}$ on set R.

The ordered pairs generated are $R= \{... (-4/3, -4/3)..(-1, -1)...(0, 0)...(3/2, 3/2)... (5, 5) ....\} $

As Identity Relation option was not given in test. I selected None but my answer was wrong.

Given options: Reflexive, Transitive, Equivalence and None.

Ubi.B
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1 Answers1

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Looking at the options given in your test the correct answer/answers are Reflexive, Transitive and Equivalence (the first two follow from the fact that $R$ is an equivalence relation). Let's check this fact:

  1. $R$ is reflexive: $xRx \iff x -x =0$ which is trivially true
  2. $R$ is symmetric: $xRy \iff yRx$ this is true because addition in $\mathbb{R}$ is a commutative operation so $x - y = 0 = y - x $
  3. $R$ is transitive: $xRy, yRz \implies xRz$ this is also true because $x -y =0\implies x=y$ and so $y-z = 0 \iff x - z = 0 \implies xRz$

So as we want this is an equivalence relation.

Turquoise Tilt
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  • I am bit confuse. Isn't this should only be identity relation as ordered pair generated are only of identity relation. E.g. (1, 1) (2.5, 2.5) etc.. then how can they be symmetric and transitive? – Ubi.B Sep 09 '22 at 09:44
  • Well what you're describing are the equivalence classes generated by $R$, indeed $(1,1)$ and $(2,2)$ aren't in the same equivalence class so they aren't in relation (this last thing could surely be written the other way round: since they aren't in relation, they aren't in the same equivalence class. I've got a question for you, could you describe how the equivalence classes of $R$ looks like? – Turquoise Tilt Sep 09 '22 at 10:10
  • For a relation to be Equivalence relation it has to be Reflexive, Symmetric and Transitive. I am confused by the ordered pairs generated by given set comprehension. – Ubi.B Sep 09 '22 at 11:03
  • The relation $R$ is an equivalence relation, we prove that earlier. Since $R$ it's a binary relation $R$ could be seen as a subset of the cartesian product $\mathbb{R} \times \mathbb{R}$ and since $xRy \iff x - y = 0 \implies x = y$ we obtain the ordered pair $(x,x)$ but this is just one element of an equivalence class. The fact is that the equivalence classes of this relation consist of single elements. Let's take for example the point $1\in \mathbb{R}$ , so $[1] = { t\in \mathbb{R} : 1Rt} = { t \in \mathbb{R} : 1 - t = 0} = {1} $. – Turquoise Tilt Sep 09 '22 at 14:03